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DENIUS [597]
1 year ago
11

A certain factory whistle can be heard up to a distance of 2.5 km. Assuming that the acoustic output of the whistle is uniform i

n all directions, how much acoustic power is emitted by the whistle? The threshold of human hearing is 1.0 × 10-12 W/m2.
Physics
1 answer:
enyata [817]1 year ago
4 0

Answer:

Emitted power will be equal to 7.85\times 10^{-5}watt

Explanation:

It is given factory whistle can be heard up to a distance of R=2.5 km = 2500 m

Threshold of human hearing I=10^{-12}W/m^2

We have to find the emitted power

Emitted power is equal to P=I\times A

P=I\times 4\pi R^2

P=10^{-12}\times 4\times 3.14\times  2500^2=7.85\times 10^{-5}watt

So emitted power will be equal to 7.85\times 10^{-5}watt

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Answer:

the ratio is \frac{V_2}{V_1}=\sqrt{2}

Explanation:

Given

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The RMS velocity of molecules in a gas is given by

V_{rms}=\sqrt{\dfrac{3k_bT}{m}}

where T=temperature

k_b=constant

For T = 387K

V_1=\sqrt{\frac{3k_b\cdot 387}{m}}----1

For T = 774

V_2=\sqrt{\frac{3k_b\cdot 774}{m}}----(2)

dividing eqn 1 and eqn 2

\frac{V_2}{V_1}=\sqrt{\frac{774}{387}}

\frac{V_2}{V_1}=\sqrt{2}

Thus,the ratio is \frac{V_2}{V_1}=\sqrt{2}

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2 years ago
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An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el
Alborosie

Answer:

I = 4.75 A

Explanation:

To find the current in the wire you use the following relation:

J=\frac{E}{\rho}      (1)

E: electric field E(t)=0.0004t2−0.0001t+0.0004

ρ: resistivity of the material = 2.75×10−8 ohm-meters

J: current density

The current density is also given by:

J=\frac{I}{A}        (2)

I: current

A: cross area of the wire = π(d/2)^2

d: diameter of the wire = 0.205 cm = 0.00205 m

You replace the equation (2) into the equation (1), and you solve for the current I:

\frac{I}{A}=\frac{E(t)}{\rho}\\\\I(t)=\frac{AE(t)}{\rho}

Next, you replace for all variables:

I(t)=\frac{\pi (d/2)^2E(t)}{\rho}\\\\I(t)=\frac{\pi(0.00205m/2)^2(0.0004t^2-0.0001t+0.0004)}{2.75*10^{-8}\Omega.m}\\\\I(t)=4.75A

hence, the current in the wire is 4.75A

4 0
1 year ago
the grid in a triode is kept negatively charged to prevent… a. the variations in voltage from getting too large. b. electrons be
Nezavi [6.7K]
D:the electrons from being attracted to the grid instead of the anode
5 0
1 year ago
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The magnetic field around a current-carrying wire is ________proportional to the current and _________proportional to the distan
PSYCHO15rus [73]

Answer:Thus, The magnetic field around a current-carrying wire is <u><em>directly</em></u>  proportional to the current and <u><em>inversely</em></u>  proportional to the distance from the wire.  If the current triples while the distance doubles, the strength of the magnetic field increases by <u><em>one and half (1.5)</em></u> times.

Explanation:

Magnetic field around a long current carrying wire is given by

B=\frac{\mu _o I}{2\pi r}

where B= magnetic field

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Thus, The magnetic field around a current-carrying wire is <u><em>directly</em></u>  proportional to the current and <u><em>inversely</em></u>  proportional to the distance from the wire.  

Now if I'=3I and r'=2r then magnetic field B' is given by

B'=\frac{\mu _oI'}{2\pi r'}=\frac{\mu _o3I}{2\pi 2r}=1.5B

Thus If the current triples while the distance doubles, the strength of the magnetic field increases by <u><em>one and half (1.5)</em></u> times.

   

7 0
2 years ago
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A flat, wide cloud floats horizontally a few kilometers above the surface of Earth. Its lower surface carries a uniform surface
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Answer:

\frac{kQ}{r^2} r^

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4 0
2 years ago
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